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Sagot :
The solution to this problem is (s2+1)Y(s)−2s−1=2s2+4 .
We let L[y(t)]=Y(s) . With that, tabular Laplace-transform laws entail
L[y′]=sY(s)−y(0)=sY(s)−2
L[y′′]=s2Y(s)−sy(0)−y′(0)=s2Y(s)−2s−1
∴L[y′′+y]=(s2+1)Y(s)−2s−1 .
Since the Laplace transform of sin(kt) is ks2+k2 , we have
(s2+1)Y(s)−2s−1=2s2+4 .
This equation readily solves to Y(s)=2s3+s2+8s+6(s2+1)(s2+4) . We compute its partial fraction decomposition as Y(s)=2s+53s2+1+−23s2+4 .
Our final step is to use tabular Laplace-transform laws ( sin(kt)↦ks2+k2 , and cos(kt)↦ss2+k2 ) to get y(t)=2cost+53sint−13sin(2t) . One can check that this indeed satisfies the initial value problem.
For more information about laplace transform, visit
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